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Saturday, May 19, 2012

Class XI, CHEMISTRY, "Chemical Equilibrium"

Reversible Reactions
Those chemical reactions which take place in both the directions and never proceed to completion are called Reversible reaction.
For these type of reaction both the forward and reverse reaction occur
at the same time so these reaction are generally represented as
Reactant □ Product
The double arrow □ indicates that the reaction is reversible and that
both the forward and reverse reaction can occur simultaneously.
Some examples of reversible reactions are given below
1. 2Hl □ H2 + l2
2. N2 + 2 H2 □ 2 NH3Irreversible Reactions
Those reactions in which reactants are completely converted into product are called Irreversible reaction.
These reaction proceed only in one direction. Examples of such type of reaction are given below
1. NaCl + AgNO3 —-> AgCl + NaNO3
2. Cu + H2SO4 —-> CuSO4 + H2Equilibrium State
The state at which the rate of forward reaction becomes equal to the rate of reverse reaction is called Equilibrium state.Explanation
Consider the following reaction
A + B □ C + D
It is a reversible reaction. In this reaction both the changes (i.e.
forward & backward) occur simultaneously. At initial stage reactant A
& B are separated from each other therefore the concentration of C
and D is zero.
When the reaction is started and the molecules of A and B react with
each other the concentration of reactant is decreased while the
concentration of product is increased. With the formation of product,
the rate of forward reaction decreased with time but the rate of reverse
reaction is increased with the formation of product C & D.
Ultimately a stage reaches when the number of reacting molecules in the
forward reaction equalizes the number of reacting molecules in the
reverse direction, so this state at which the rate of forward reaction
becomes equal to the rate of reverse reaction is called equilibrium
state.Law of Mass ActionStatement
The rate at which a substance reacts is proportional to its active mass
and the rate of a chemical reaction is proportional to the product of
the active masses of the reactant.
The term “active mass” means the concentration in terms of moles/dm3.Derivation of Equilibrium Constant Expression
Consider in a reversible reaction “m” mole of A and “n” moles of B
reacts to give “x” moles of C and “y” moles of D as shown in equation.
mA + nB □ xC + yD
In this process
The rate of forward reaction ∞ [A]m [B]n
Or
The rate of forward reactin = Kf [A]m [B]n
&
The rate of reverse reaction ∞ [C]x [D]y
Or
The rate of reverse reaction = Kf [C]x [D]y
But at equilibrium state
Rate of forward reaction = Rate of reverse reaction
Therefore,
Kf [A]m [B]n = Kf [C]x [D]y
Or
Kf / Kr = [C]x [D]y / [A]m [B]n
Or
Ke = [C]x [D]y / [A]m [B]n
This is the expression for equilibrium constant which is denoted by Ke and defined as
The ratio of multiplication of active masses of the products to the
product of active masses of reactant is called equilibrium constant.Equilibrium Constant for a Gaseous System
Consider in a reversible process, the reactants and product are gases as shown
A(g) + B(g) □ C(g) + D(g)
When the reactants and products are in gaseous state, their partial
pressures are used instead of their concentration, so according to law
of mass action.Determination of Equilibrium Constant
The value of equilibrium constant K(C) does not depend upon the initial
concentration of reactants. In order to find out the value of K(C) we
have to find out the equilibrium concentration of reactant and product.1. Ethyl Acetate Equilibrium
Acetic acid reacts with ethyl alcohol to form ethyl acetate and water as shown
CH3COOH + C2H5OH □ CH3COOC2H5 + H2O
Suppose ‘a’ moles of acetic acid and ‘b’ moles of alcohol are mixed in
this reaction. After some time when the state of equilibrium is
established suppose ‘x’ moles of H2O and ‘x’ moles of ethyl acetate are
formed while the number of moles of acetic acid and alcohol are a-x and
b-x respectively at equilibrium.
According to law of mass action
K(C) = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
K(C) = [x/V] [x/V] / [a-x/V] [b-x/V]
K(C) = (x) (x) / (a-x) (b-x)
K(C) = x2 / (a-x) (b-x)2. Hydrogen Iodide Equilibrium
For the reaction between hydrogen and iodine suppose a mole of hydrogen
and ‘b’ moles of iodine are mixed in a scaled bulb at 444ºC in the
boiling sulphur for some time. The equilibrium mixture is then cooled
and the bulbs are opened in the solution of NaOH. Let the amount of
hydrogen consumed at equilibrium be ‘x’ moles which means that the
amount of hydrogen left at equilibrium is a-x moles. Since 1 mole of
hydrogen reacts with 1 mole of iodine ‘o’ form two moles of hydrogen
iodide hence the amount of iodine used is also x moles so its moles at
equilibrium are b-x and the moles of hydrogen iodide at equilibrium are
2x.
According to law of mass action
K(C) = [Hl]2 / [H2] [l2]
K(C) = [2x/V]2 / [a-x/V] [b-x/V]
K(C) = 4×2 / (a-x) (b-x)Applications of Law of Mass Action
There are two important applications of equilibrium constant.
1. It is used to predict the direction of reaction.
2. K(C) is also used to predict the extent of reaction.To Predict the Direction of Reaction
The value of equilibrium constant K(C) is used to predict the direction of reaction. For a reversible process.
Reactant □ Product
With respect to the ratio of initial concentration of the reagent.
There are three possibilities for the value of K
1. It is greater than K(C)
2. It is less than K(C)
3. It is equal to K(C)Case I
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the reverse direction.Case II
If [Reactant]initial / [Product]initial > K(C) the reaction will shift towards the forward direction.Case III
If [Reactant]initial / [Product]initial > K(C) this is equilibrium state for the reaction.To Predict the Extent of Reaction
From the value of K(C) we can predict the extent of the reaction.
If the value of K(C) is very large e.g.
For 2 O3 □ 3 O2 ……….. K(C) = 10(55)
From this large value of K(C) it is predicted that the forward reaction is almost complete.
When the value of K(C) is very low e.g.,
2 HF □ H2 + F2 ……….. K(C) = 10(-13)
From this value it is predicted that the forward reaction proceeds with negligible speed.
But if the value of K(C) is moderate, the reaction occurs in both the
direction and equilibrium will be attained after certain period of time
e.g., K(C) for
N2 + 3 H2 □ 2 NH3 …………. is 10
So the reaction occurs in both the direction.Le Chatelier’s PrincipleStatement
When a stress is applied to a system at equilibrium the equilibrium
position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing
concentration pressure or temperature. The effect of these changes is
explained by Le Chatelier.Effect of Concentration
By changing the concentration of any substance present in the
equilibrium mixture, the balance of chemical equilibrium is disturbed.
For the reaction,
A + B □ C + D
K(C) = [C][D] / [A][B]
If the concentration of a reactant A or B is increased the equilibrium state shifts tc right and yield of products increases.
But if the concentration of C or D is increased then the reaction
proceed in the backward direction with a greater rate and more A & B
are formed.Effect of Temperature
The effect of temperature is different for different type of reaction.
For an exothermic reaction the value of K(C) decreased with the increase
of temperature so the concentration of products decreases.
For a endothermic reaction heat is absorbed for the conversion of
reactant into product so if temperature during the reaction is increased
then the reaction will proceed with a greater rate in forward
direction.ENDOTHERMIC REACTION
Temperature increase —-> More products are formed
Temperature decrease —-> More reactants are formedEXOTHERMIC REACTION
Temperature increase —-> More reactants are formed
Temperature decrease —-> More products are formedEffect of Pressure
The state of equilibrium of gaseous reaction is distributed by the
change of pressure. There are three types of reactions which show the
effect of pressure change.1. When the Number of Moles of Product are Greater
In a reaction such as
PCl5 <—-> PCl3 + Cl2
The increase of pressure shifts the equilibrium towards reactant side.2. When the Number of Moles of Reactant are Greater
In a reaction such as
N2 + 3H2 <—-> 2NH3
The increase of pressure shifts the equilibrium towards product side
because the no. of moles of product are less than the no. of moles of
reactant.3. When Number of Moles of Reactants and Products are Equal
In these reactions where the number of moles of reactant are equal to
the number of moles of product the change of pressure does not change
the equilibrium state e.g.,
H2 + l2 □ 2 Hl
Since the number of moles of reactants and products are equal in this
reaction so the increase of pressure does not affect the yield of Hl.Important Industrial Application of Le Chatelier’s PrincipleHaber’s Process
This process is used for the production of NH3 by the reaction of
nitrogen and hydrogen. In this process 1 volume of nitrogen is mixed
with three volumes of hydrogen at 500ºC and 200 to 1000 atm pressure in
presence of a catalyst
N2 + 3 H2 □ 2 NH3 …………… ΔH = -46.2 kJ/mole1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [NH3]2 / [N2] [H2]3
Increase in concentration of reactants which are nitrogen and hydrogen
the equilibrium of the process shifts towards the right so as to keep
the value of K(C) constant. Hence the formation of NH3 increases with
the increase of the concentration of N2 or hydrogen.2. Effect of Temperature
It is an exothermic process, so heat is liberated with the formation of
product. Therefore, according to Le Chatelier’s principle at low
temperature the equilibrium shifts towards right to balance the
equilibrium state so low temperature favours the formation of NH33. Effect of Pressure
The formation of NH3 proceeds with the decrease in volume, therefore,
the reaction is carried out under high pressure or in other words high
pressure is favourable for the production of NH3.Contact Process
The process is used to manufacture H2SO4 on large scale. In this
process the most important step is the oxidation of SO2 to SO3 in
presence of a catalyst vanadium pentoxide.
2 SO2 + O2 □ 2 SO3 ………………. ΔH = – 395 kJ/mole1. Effect of Concentration
The value of K(C) for this reaction is
K(C) = [SO3]2 / [SO2]2 [O2]
Increase in concentration of SO2 or O2 shifts the equilibrium towards the right and more SO3 is formed.2. Effect of Temperature
Since the process is exothermic, so low temperature will favour the
formation of SO3. The optimum temperature for this reaction is 400 to
450ºC.3. Effect of Pressure
In this reaction decrease in volume takes place so high pressure is favourable for the formation of SO3.Common Ion EffectStatement
The process in which precipitation of an electrolyte is caused by
lowering the degree of ionization of a weak electrolyte when a common
ion is added is known as common ion effect.Explanation
In the solution of an electrolyte in water, there exist an equilibrium
between the ions and the undissociated molecules to which the law of
mass action can be applied.
Considering the dissociation of an electrolyte AB we have
AB □ A+ + B-
And
[A+][B-] / [AB] = K (dissociation constant)
If now another electrolyte yielding A+ or B- ions be added to the above
solution, it will result in the increase of concentration of the ions A+
or B- and in order that K may remain the same, the concentration AB
must evidently increase. In other words the degree of dissociation of an
electrolyte is suppressed by the addition of another electrolyte
containing a common ion. This phenomenon is known as common ion effect.Application of Common Ion Effect in Salt Analysis
An electrolyte is precipitated from its solution only when the
concentration of its ions exceed from the solubility product. The
precipitates are obtained when the concentration of any one ion is
increased. Thus by adding the common ion, the solubility product can be
exceeded.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
When Na2CO3 is dissolved in water, it reacts with water such as
Na2CO3 + 2 H2O □ 2 NaOH + H2CO3
In this solution H2CO3 which is weak acid an NaOH which is a strong base
are formed. Due to presence of strong base the medium is changed
towards basic nature.Solubility Product
When a slightly soluble ionic solid such as silver chloride is dissolved in water, it decompose into its ions
AgCl □ Ag+ + Cl-
These Ag+ and Cl- ions from solid phase pass into solution till the
solution becomes saturated. Now there exists an equilibrium between the
ions present in the saturated solution and the ions present in the solid
phase, thus
AgCl □ Ag+ + Cl-
Applying the law of mass action
K(C) = [Ag+][Cl-] / [AgCl]
Since the concentration of solid AgCl in the solid phase is fixed, no
matter how much solid is present in contact with solution, so we can
write.
K(C) = [Ag+][Cl-] / K
Or
K(C) x K = [Ag+][Cl-]
Or
K(S.P) = [Ag+][Cl-]
Where K(S.P) is known as solubility product and defined as
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is called solubility product.
the value of solubility product is constant for a given temperature.Calculation of Solubility Product From Solubility
The mass of a solute present in a saturated solution with a fixed volume
of solvent is called solubility, which is generally represented in the
unit of gm/dm3. With the help of solubility we can calculate the
solubility product of a substance e.g., the solubility of Mg(OH)2 at
25ºC is 0.00764 gm/dm3. To calculate the K(S.P) of Mg(OH)2, first of all
we will calculate the concentration of Mg(OH)2 present in the solution.
Mass of Mg(OH)2 = 0.00764 gm/dm3
Moles of Mg(OH)2 = 0.00764 / 58 moles / dm3
= 1.31 x 10(-4) moles/dm3
The ionization of Mg(OH)2 in the solution is as follows.
Mg(OH)2 □ Mg(+2) + 2 OH-
And the solubility product for Mg(OH)2 may be written as,
K(S.P) = [Mg(+2)] [OH-]2
Since in one mole of Mg(OH2) solution one mole of Mg++ ions are present
while two moles of OH- ions are present, therefore in 1.31 x 10(-4)
mole/dm3 solution of Mg(OH)2, the concentration of Mg(+2) is 1.31 x
10(-4) moles/dm3 while the concentration of OH- is 2. 62 x 10(-8)
moles/dm3. By substituting these values
K(S.P) = [Mg(+2)][OH-]2
= [1.31 x 10(-4)] [2.62 x 10(-4)]2
= 9.0 x 10(-12) mole3 / dm9
So in this way the solubility product of a substance may be calculated with the help of solubility.Calculation of Solubility from Solubility Product
If we know the value of solubility product, we can calculate the solubility of the salt.
For example, the solubility of PbCrO4at 25ºC is 2.8 x 10(-13) moles/dm3.
m = n2 / w1 in kg
m = (w2 / m2) / (w1 / 1000)
m = w2 / m2) x (1000 / w1)Hydration
Addition of water or association of water molecules with a substance without dissociation is called Hydration.
Water is a good solvent and its polar nature plays very important part
in dissolving substances. It dissolves ionic compounds readily.
When an ionic compound is dissolved in water, the partial negatively
charged oxygen of water molecule is attracted towards the cation ion
similarly the partial positively charged hydrogen of water molecule is
attracted towards the anions so hydrated ions are formed.
Diagram Coming Soon
In solution, the number of water molecules which surround the ions is
indefinite, but when an aqueous solution of a salt is evaporated the
salt crystallizes with a definite number of water molecules which is
called as water of crystallization E.g., when CuSO4 recrystallized from
its solution the crystallized salt has the composition CuSO4. 5H2O.
Similarly when magnesium chloride is recrystallized from the solution,
it has the composition MgCl2.6H2O. This composition indicates that each
magnesium ion in the crystal is surrounded by six molecules. This type
of salts is called hydrated salts.
It is observed experimentally that the oxygen atom of water molecule is
attached with the cation of salt through co-ordinate covalent bond so it
is more better to write the molecular formulas of the hydrated salts as
given below.
[Cu(H2O)5]SO4 …………….. [Mg(H2O)6]Cl2
It is also observed that these compound exist with a definite
geometrical structure e.g., the structure of [Mg(H2O)6]Cl2 is octahedral
and [Cu(H2O)4]+2 is a square planar.
Diagram Coming SoonFactors for Hydration
The ability of hydration of an ion depend upon its charge density.
For example the charge density of Na+ is greater than K+ because of its
smaller size, so the ability of hydration for Na+ is greater than K+
ion. Similarly small positive ions with multiple charges such as Cu(+2),
Al(+3), Cr(+3) posses great attraction for water molecules.Hydrolysis
Addition of water with a substance with dissociation into ions is called Hydrolysis.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoritically it is expected that the solution of salts like CuSO4 or
Na2CO3 are neutral because these solutions contain neither H+ ion nor
OH-, but it is experimentally observed that the solution of CuSO4 is
acidic while the solution of Na2CO3 is basic. This acidic or basic
nature of solution indicate but H+ ions or OH- ions are present in their
solutions which can be produced only by the dissociation of water
molecules.Theory of Ionization
1n 1880, a Swedish chemist Svante August Arrhenius put forward a
theory known as theory of ionization, in order to account for the
conductivity of electrolytes, electrolysis and certain properties of
electrolytic solutions. According to this theory.
1. Acids, Bases and Salts when dissolved in water yield two kinds of
ions, one carry positive charge and the other carry negative charge. The
positively charged ions are called cations which are derived from
metals or it may be H+ ion but the negatively charged ions which are
known as anions are derived from non-metals
NaCl —-> Na+ + Cl-
H2SO4 —-> 2 H+ + SO4(-2)
KOH —-> K+ + OH-
2. Ions in the solution also recombine with each other to form
neutral molecules and this process continues till an equilibrium state
between an ionized and unionized solid is attained